SiNx = 1 / 2, and X ∈ [0,2 π], then x=

SiNx = 1 / 2, and X ∈ [0,2 π], then x=

30 ° or 150 degree

SiNx ＜ X. if x = 90 degrees, then SiNx = 1 ＜ 90 degrees? Here,
SiNx = 1 ＜ 90 degrees, it seems that there is no relationship between them! If it does, then 90 degrees =?
Is x a number or an angle

Here, X represents an angle in radians, so 90 degrees is x = π / 2
sinπ/2=1
x=π/2≈3.14/2=1.57
That is, when x = π / 2, substituting SiNx ＜ x, we get the following result:
sinπ/2

SiNx = 1 / 2, and X ∈ [0,2 π], then x=
The answer is π / 6, 5 π / 6. Its value range is 0 ~ 360 degrees. Why is it 5 π / 6

Hello, first SiNx > 0
It shows that x is in quadrant 1,2
5π/6=150°
sin30°=1/2
Induction formula
sin30°=sin（180°-30°)=sin150°=1/2
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It is proved that sinx-x = 1 has only one root between - 2 and - 1

f(x)=sinx-x-1
f'(x)=cosx-1
So f '(x) 0
f(-1)=-sin1-1+1=-sin1

limx→∞ (x^2-x+1)(3-sinx)/x^3+2

lim (x^2-x+1)(3-sinx)/(x^3+2)
= lim (1-1/x+1/x^2)(3-sinx)/(x+2/x^2) = 0

Using cosx = sin (π / 2-x), sin'x = cosx, it is proved that (cosx) '= - SiNx

(cosx)'=[sin(π/2-x)]‘=cos(π/2-x)*(π/2-x)'=sinx*(-1)=-sinx

The maximum and minimum values of the function f (x) = xcosx SiNx on (0,2 π) are?

f'(x)=cosx-xsinx-cosx=-xsinx
Because 0

What is the maximum and minimum of the function f (x) = xcosx SiNx on [0,2 π]?

F (x) = xcosx sinxf '(x) = - xsinx + cosx - cosx = 0xsinx = 0 or 2 π f' '(x) = - (xcosx + SiNx) f' '(0) = 0f' '(2 π) = 2 π > 0 (min) minf (x) = f (2 π) = - 2 π X in (0,2 π) f' (x)

Let f (x) = SiNx xcosx, X be all real numbers. When x > 0, the monotone interval of the function. If x belongs to [02013 π], the sum of all extremums is obtained

f'(x)=cosx-(cosx-xsinx)=xsinx
When x > 0, SiNx = 0, x = k, π is the extreme point, and k > 1 is an integer
Monotone increasing interval (0, π / 2) U (2k π - π / 2,2k π + π / 2), k = 1,2
Monotone decreasing interval (2k π + π / 2,2k π + 3 π / 2), k = 0,1,2
The extreme points are x = k π, k = 0,1,2,..., 2013
f(kπ)=-kπ(-1)^k
So the sum of extremum = - π [0-1 + 2-3 +, + 2012-2013] = - π (- 1007) = 1007 π

F (x) = m SiNx + n xcosx + x ^ 2, and f (3) = 10, f (- 3) =?

f(x)=m sinx+n xcosx+x^2
f(3)=10
f(3)=m sin3+n 3cos3+3^2
f(-3)=m sin(-3)+n(-3)cos(-3)+(-3)^2
=-m sin3-n 3cos3+3^2
=-m sin3-n 3cos3-3^2+2*3^2
=-f(3)+2*3^2
=-10+18=8